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oee [108]
3 years ago
13

How much will Chris’s investment account be worth in five years if he invests $5,000 with 2.8% interest?

Mathematics
1 answer:
vovikov84 [41]3 years ago
6 0

Answer:

\$5,700

Step-by-step explanation:

we know that

The simple interest formula is equal to

A=P(1+rt)

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

t=5\ years\\ P=\$5,000\\r=0.028

substitute in the formula above

A=\$5,000(1+0.028*5)

A=\$5,000(1.14)=\$5,700

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You have 5 Red Marbles, 4 Green Marbles, and 3 Blue Marbles in a Bowl. You randomly pick 4 marbles from the bowl (without replac
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Answer:

2%

Step-by-step explanation:

In this case each one is an independent event, therefore, the multiplication of each one would be the final probability

We have 5 Red Marbles, 4 Green Marbles, and 3 Blue Marbles, that is, there are 5 + 4 + 3 12 Marbles in total.

Now if I draw a red one, the probability would be: 5/12

When drawing another red, the probability would be: 4/11

When taking the green: 4/10

When removing the blue: 3/9

Finally, the final probability is:

P = (5/12) * (4/11) * (4/10) * (3/9)

P = 0.020

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3 years ago
A random survey of 1000 students nationwide showed a mean ACT score of 21.1. Ohio was not used. A survey of 500 randomly selecte
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Answer:

We conclude that the mean Ohio score is below the national average.

Step-by-step explanation:

We are given that a random survey of 1000 students nationwide showed a mean ACT score of 21.1. Ohio was not used.

A survey of 500 randomly selected Ohio scores showed a mean of 20.8. The population standard deviation is 3.

<u><em>Let </em></u>\mu<u><em> = mean Ohio scores.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 21.1      {means that the mean Ohio score is above or equal the national average}

Alternate Hypothesis, H_A : \mu < 21.1      {means that the mean Ohio score is below the national average}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about population standard deviation;

                    T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n}}}  ~ N(0,1)

where, \bar X = sample mean Ohio score = 20.8

            \sigma = population standard deviation = 3

            n = sample of Ohio = 500

So, <u><em>test statistics</em></u>  =  \frac{20.8-21.1}{\frac{3}{\sqrt{500}}}  

                              =  -2.24

The value of z test statistics is -2.24.

<em>Now, at 0.1 significance level the z table gives critical value of -1.2816 for left-tailed test.</em><em> Since our test statistics is less than the critical values of z as -2.24 < 1.2816, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the mean Ohio score is below the national average.

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